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Inverse/Reverse of Number of Permutations and of Number of Combinations with Repetitions?

0

$begingroup$

For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.

In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...

Given:

$X = C(n, k) = binomnk = fracn! k!(n-k)! $

then you can limit n to:

$sqrt[k]k! X + k ge n ge sqrt[k]k! X$

And thus you at most need to check k+1 possible values of n.
Great!

But how do I solve for k instead of for n?

And given instead:

$X = P(n, k) = fracn! (n-k)! $

then would I be correct in assuming from above that:

$sqrt[k]X + k ge n ge sqrt[k]X$

? And if so, how would I then solve for k instead of n?

And finally, given instead CR is Combinations with Repetitions:

$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $

then would I be correct in assuming from above that:

$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$

? And if so, how would I then solve for k instead of n?

Any help on the algebra for these five related inverse functions would be greatly appreciated!

combinatorics permutations combinations binomial-coefficients inverse-function

share|cite|improve this question

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Question: when we are solving for $k$, is $n$ fixed?
  $endgroup$
  – Eevee Trainer
  Mar 27 at 7:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
  $endgroup$
  – Brian Kennedy
  Mar 27 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
  $endgroup$
  – Mike Earnest
  Mar 28 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
  $endgroup$
  – Mike Earnest
  Mar 28 at 16:33
  

  
  
  
  

add a comment | 

0

$begingroup$

For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.

In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...

Given:

$X = C(n, k) = binomnk = fracn! k!(n-k)! $

then you can limit n to:

$sqrt[k]k! X + k ge n ge sqrt[k]k! X$

And thus you at most need to check k+1 possible values of n.
Great!

But how do I solve for k instead of for n?

And given instead:

$X = P(n, k) = fracn! (n-k)! $

then would I be correct in assuming from above that:

$sqrt[k]X + k ge n ge sqrt[k]X$

? And if so, how would I then solve for k instead of n?

And finally, given instead CR is Combinations with Repetitions:

$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $

then would I be correct in assuming from above that:

$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$

? And if so, how would I then solve for k instead of n?

Any help on the algebra for these five related inverse functions would be greatly appreciated!

combinatorics permutations combinations binomial-coefficients inverse-function

share|cite|improve this question

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Question: when we are solving for $k$, is $n$ fixed?
  $endgroup$
  – Eevee Trainer
  Mar 27 at 7:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
  $endgroup$
  – Brian Kennedy
  Mar 27 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
  $endgroup$
  – Mike Earnest
  Mar 28 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
  $endgroup$
  – Mike Earnest
  Mar 28 at 16:33
  

  
  
  
  

add a comment | 

$begingroup$

For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.

In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...

Given:

$X = C(n, k) = binomnk = fracn! k!(n-k)! $

then you can limit n to:

$sqrt[k]k! X + k ge n ge sqrt[k]k! X$

And thus you at most need to check k+1 possible values of n.
Great!

But how do I solve for k instead of for n?

And given instead:

$X = P(n, k) = fracn! (n-k)! $

then would I be correct in assuming from above that:

$sqrt[k]X + k ge n ge sqrt[k]X$

? And if so, how would I then solve for k instead of n?

And finally, given instead CR is Combinations with Repetitions:

$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $

then would I be correct in assuming from above that:

$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$

? And if so, how would I then solve for k instead of n?

Any help on the algebra for these five related inverse functions would be greatly appreciated!

combinatorics permutations combinations binomial-coefficients inverse-function

share|cite|improve this question

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062

$endgroup$

share|cite|improve this question

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062

share|cite|improve this question

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062

asked Mar 27 at 6:52

Brian KennedyBrian Kennedy

1062